Path: utzoo!utgpu!news-server.csri.toronto.edu!mailrus!wuarchive!zaphod.mps.ohio-state.edu!samsung!crackers!cpoint!frog!john
From: john@frog.UUCP (John Woods)
Newsgroups: sci.electronics
Subject: Re: 12V to 5V conversion
Message-ID: <16640@frog.UUCP>
Date: 13 Jul 90 18:59:00 GMT
References: <987@cameron.egr.duke.edu>
Distribution: usa
Organization: Misanthropes-R-Us
Lines: 23

In article <987@cameron.egr.duke.edu>, wdp@dukee.egr.duke.edu (Dev Palmer) writes:
> Since you are dropping 7 V across the regulator, you are generating quite a 
> bit of heat and the regulators may shut down or at least begin to shut down.
> If you really want 3.00 A (giving 15.00 W),... use the LM350K adjusted to
> 5 V and mounted on a BIG heat sink

This would be my recommendation, too, but another thing to try would be a
resistor in series between the battery and regulator.  Set it to drop 4-5
volts at maximum current (1.3 ohms or so at 3A), and it will dissipate half
the power for you, keeping the regulator(s) much cooler.  Keep the resistors
away from the electronics, and everything runs cooler.  I use a similar
trick on this computer (frog) to run a 12V disk drive from a 24V supply.
If your heat sinking is really effective, you can get by with the cheaper
LM350T (plastic TO220) but in frog's case, it helps that there is a fan
1/2 inch from the heatsink!

The LM350 is a great part.  The 5A and 10A regulators are so much more
expensive that for just screwing around it makes more sense to wire a
pass transistor by hand, but the LM350 is cheap enough that it's better to
use it instead.
-- 
John Woods, Charles River Data Systems, Framingham MA, (508) 626-1101
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